## Coursera: analysis complex kind Week 3.1. Differentiability, continuity, analytic functions

### Jul 17, 2016

Key definitions in this post.

### Slide 2

Reviewed the definition of real derivative of f(x) at $x_0$ as limit

$\underset{x \to {x_0}}{\lim} \, \frac{f(x) - f(x_0)}{x - x_0}$

### Slides 3 & 4

Graphical examples given as secant becomes tangent.

### Slide 5

“Failing” Example when graph has a “point”, that is the gradient of the curve is different each side so gradient limit from either side of x0 is not the same, therefore derivative at x0 does not exist.

### Slide 6

Complex derivative is essentially the same

$\underset{z \to {z_0}}{\lim} \, \frac{f(z) - f(z_0)}{z - z_0}$

… different conventions that represent derivative here ..

or, if h is a small complex number, and we write z as $z_0 + h$ the limit can be written as

$\underset{h \to 0}{\lim} \, \frac{f({z_0}+h) - f(z_0)}{h}$

Since h is any complex number, $z_0$ can be “approached” from any  “direction”.

This definition of the complex derivative as a limit with same “form” as real derivative leads to (?) the same standard results as in real-valued calculus

e.g. d/dz(c f(z)) = c f ‘(z), c is a constant

• product rule applies
• chain rule applies

as a result complicated functions, e.g. $z^2 e^{sin z}$ can sometimes be split into simpler components for differentiation.

## *Important* Slide 11 – non examples

Consider f(z) = Re(z)

If z = x + iy, so f(z) = Re(z) = x and let h = $h_x + i h_y$

$\frac{{f({z}}+{h}) - f(z)}{h} = \frac{(x+h_x) - x}{h}= \frac{h_x}{h}= \frac{Re(h)}{h}$ (*)

So .. does this have a limit as h -> 0?

h-> 0 along real axis. Then $h = h_x + i0 = h_x$

Substituting into (*) we get h / h = 1

Whereas, h-> 0 along imaginary axis. Then $h = 0 + ih_y = i h_y$

Since Re(h) now equals zero, when we substitute this into (*) we get a quotient value of zero.

So depending on how h approaches zero (along real or imaginary axes) the difference quotient has two values, 1 and zero, so f is not differentiable at z = 0.

Further …

Consider h to be a sequence of values that spiral into towards the origin but alternating between real and imaginary values

<< sketch here would help >>

A suitable function (which we put in place of h) could be

$h_n = \frac{i^n}{n}$, where n is a positive integer, creating the sequence

$\frac{i^1}{1},\frac{i^2}{2},\frac{i^3}{3},\frac{i^4}{4},\frac{i^5}{5}, ... = i, \frac{-1}{2},\frac{-i}{3},\frac{1}{4},\frac{i}{5}, ...$

So that sequence of h values is spiralling towards zero

And the difference quotient (*) becomes $\frac{Re \; h_n}{h_n} \; = \; \frac{Re \; i^n}{i^n}$

Which has the values

1 when n is even: $\frac{ Re \; i^2}{i^2}, \frac{Re \; i^4}{i^4}, ... = \frac{-1}{-1}, \frac{1}{1},, ...$

0 when n is odd $\frac{ Re \; i^1}{i^1},\frac{Re \; i^3}{i^3},\frac{Re \; i^5}{i^5}, ... = 0, \frac{0}{-i},\frac{0}{i}, ...$

so the sequence has no limit, and since z was arbitrary, function is not differentiable anywhere on C.

## Slide 12 Another non-example f(z) = $\overline{z}$

Let f(z) = $\overline{z}$

So defining z and h as in slide 11

$\frac{f({z}+h) - f(z)}{h} = \frac{(\overline{z}+\overline{h}) - (\overline{z})}{h}= \frac{\overline{h}}{h}$

If h $\in$ R then $\frac{\overline{h}}{h}$ = 1 -> 1 as h -> 0.

If h $\in$ iR then $\frac{\overline{h}}{h}$ = -1 -> -1 as h -> 0.

Thus $\frac{\overline{h}}{h}$ does NOT have a limit as h -> 0, and since z is arbitrary, f (complement of z) is not differentiable anywhere in C.

## ** Slide 14 Analytic functions **

Definition : A function is analytic in an open subset U ⊂ C, if it is complex differentiable at every point in U

Definition: A function which is analytic on all of C is entire.

Examples:

Polynomials analytic every in C therefore entire

Rational functions p(x)/q(x) are analytic when q(x) not equal zero.

Modulus function not analytic

Re(z), Im(z) not analytic

Another example, by first principles f(z) =( modulus(z))^2 is differentiable only at the single point z=0 therefore not analytic, however it is continuous everywhere in C

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